园丁计划建造两个花坛，一个是半径为 $r$ m 的圆，另一个是边长为 $x$ m 的正方形。为了保护幼花，这两个花坛将用铁丝网围起来。所用铁丝网的总长度为 30 m。

1. Formulate an equation for the total length of the wire netting

The perimeter of the circle is $2\pi r$ and the perimeter of the square is $4x$.
Total length = $2\pi r + 4x = 30$

2. Make $x$ the subject of the formula

$4x = 30 - 2\pi r$
$x = \frac{30 - 2\pi r}{4}$
$x = \frac{15 - \pi r}{2}$

1. State the expression for the combined area in terms of $r$ and $x$, and recall $x$ from part (i)

From part (i), the total perimeter is $2\pi r + 4x = 30$, so $x = \frac{15 - \pi r}{2}$.
The combined area $A$ is the sum of the area of the circle and the area of the square:
$A = \pi r^2 + x^2$

2. Substitute $x$ into the area equation and expand

$A = \pi r^2 + \left(\frac{15 - \pi r}{2}\right)^2$
$A = \pi r^2 + \frac{225 - 30\pi r + \pi^2 r^2}{4}$

3. Simplify and group terms to match the required form

$A = \pi r^2 + \frac{225}{4} - \frac{30\pi r}{4} + \frac{\pi^2 r^2}{4}$
$A = \pi r^2 + \frac{1}{4}\pi^2 r^2 - \frac{15}{2}\pi r + \frac{225}{4}$
$A = \left(\frac{1}{4}\pi^2 + \pi\right)r^2 - \frac{15}{2}\pi r + \frac{225}{4}$

1. Differentiate the expression for $A$ with respect to $r$.

Given $A = \left(\frac{1}{4}\pi^2 + \pi\right)r^2 - \frac{15}{2}\pi r + \frac{225}{4}$,
$\frac{\mathrm{d}A}{\mathrm{d}r} = 2\left(\frac{1}{4}\pi^2 + \pi\right)r - \frac{15}{2}\pi$
$\frac{\mathrm{d}A}{\mathrm{d}r} = \left(\frac{1}{2}\pi^2 + 2\pi\right)r - \frac{15}{2}\pi$

2. Set the derivative to zero to find the stationary value.

For a stationary value, $\frac{\mathrm{d}A}{\mathrm{d}r} = 0$.
$\left(\frac{1}{2}\pi^2 + 2\pi\right)r - \frac{15}{2}\pi = 0$

3. Solve for $r$.

$\left(\frac{1}{2}\pi^2 + 2\pi\right)r = \frac{15}{2}\pi$
Multiply by 2:
$(\pi^2 + 4\pi)r = 15\pi$
$r = \frac{15\pi}{\pi^2 + 4\pi}$
$r = \frac{15}{\pi + 4}$

1. Recall the expression for $x$ from part (i) and the stationary value of $r$ from part (iii).

From (i), $x = \frac{15 - \pi r}{2}$.
From (iii), the stationary value occurs when $r = \frac{15}{\pi + 4}$.

2. Substitute $r$ into the expression for $x$ and simplify to show it equals $2r$.

$x = \frac{15 - \pi\left(\frac{15}{\pi + 4}\right)}{2}$
$x = \frac{\frac{15(\pi + 4) - 15\pi}{\pi + 4}}{2}$
$x = \frac{15\pi + 60 - 15\pi}{2(\pi + 4)}$
$x = \frac{60}{2(\pi + 4)} = \frac{30}{\pi + 4}$
Since $r = \frac{15}{\pi + 4}$, the diameter of the circle is $2r = \frac{30}{\pi + 4}$.
Therefore, $x = 2r$ (shown).

1. Find the second derivative of $A$ with respect to $r$

From part (ii), $A = \left(\frac{1}{4}\pi^2 + \pi\right)r^2 - \frac{15}{2}\pi r + \frac{225}{4}$.

$\frac{dA}{dr} = 2\left(\frac{1}{4}\pi^2 + \pi\right)r - \frac{15}{2}\pi$

$\frac{d^2A}{dr^2} = 2\left(\frac{1}{4}\pi^2 + \pi\right) = \frac{1}{2}\pi^2 + 2\pi$

2. Determine the nature of the stationary value using the sign of the second derivative

Since $\pi > 0$, $\frac{d^2A}{dr^2} = \frac{1}{2}\pi^2 + 2\pi > 0$.

Therefore, the stationary value of $A$ is a minimum.